Why does the natural ring homomorphism induce a surjective group homomorphism of units? I'm trying an old problem here: http://www.math.dartmouth.edu/archive/m111s09/public_html/homework-posted/hw1.pdf
Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?
My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?
 A: First reduce to the case where $n$ and $m$ are both powers of a prime $p$, using the Chinese remainder theorem. After this reduction the result is clear since the $p$-adic valuation of $n$ is lesser than the $p$-adic valuation of $m$ for any prime $p$.
A: If you want a "pedestrian" solution without knowledge in group theory or number theory, you have to note that when $k$ is your unit, then $k+ln$ are also units, so you have several candidates for the unit in the bigger group and you only have to plug them into your equation and show that one of them gives a coefficient of $n$ divisible by $a$.
A: This answer is just to observe that this is a special case of a very general result: if $R$ is an artinian ring and $I$ is a two-sided ideal of $R$, then the quotient map induces a surjection $R^\times \to (R/I)^\times$ (finite rings such as $\mathbf{Z}/n \mathbf{Z}$ are, of course, artinian). There is also closely related fact that for any ring $R$, the quotient map $R^\times \to (R/J)^\times$ is surjective, where $J$ is the Jacobson radical of $R$.
